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How do I solve this equation:

$$ 2\arcsin\frac{x}{2}+\arcsin(x\sqrt{2})=\frac{\pi}{2} $$

We know that:

$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$

So letting $\alpha = 2\arcsin\frac{x}{2}$ and $\beta=\arcsin(x\sqrt{2})$ leads to: $\sin\frac{\alpha}{2} = \frac{x}{2}$ and $\sin\beta = x\sqrt{2}$. Finding $\sin\alpha$ and $\cos\alpha$ first:

$$\begin{align} \cos \frac{\alpha}{2} & = \frac{\sqrt{(4-x^2)}}{2} \\ \sin\alpha & = \sin\left(\frac{\alpha}{2} + \frac{\alpha}{2}\right) = 2\cos\frac{\alpha}{2}\sin\frac{\alpha}{2}\\ & = \frac{x\sqrt{(4-x^2)}}{2} \\ \cos \alpha & = \frac{\sqrt{(4 - (x\sqrt{(4-x^2)})^2)}}{2} = \frac{\sqrt{(4 - x^2(4-x^2))}}{2} \end{align}$$

And now $\cos\beta$:

$$\begin{align} \sin \beta & = x\sqrt2 \\ \cos \beta & = \sqrt{1 - 2x^2} \end{align}$$

Plugging everything together:

$$ 1 = \frac{x\sqrt{(4-x^2)} \times \sqrt{1 - 2x^2}}{2} + \frac{\sqrt{(4 - x^2(4-x^2))} \times x\sqrt2}{2} \\ 2 = x\sqrt{4-9x^2+2x^4} + x\sqrt{8-8x^2+2x^4} \\ 4 = x^2(4-9x^2+2x^4) + x^2(8-8x^2+2x^4) \\ 0 = 4x^4 -17x^3 + 12x^2 - 4 $$

Which is incorrect - the correct answer is $\sqrt{6-4\sqrt2}$. Where did I go wrong?

hohner
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1 Answers1

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Let $\arcsin\dfrac x2=y\implies x=2\sin y$

So we have, $$\arcsin(\sqrt2\cdot2\sin y)=\dfrac\pi2-2y$$

Applying sine on both sides, $$\sin\left[\arcsin(\sqrt2\cdot2\sin y)\right]=\sin\left(\dfrac\pi2-2y\right)=\cos2y$$

$$\implies\sqrt2\cdot2\sin y=1-2\sin^2y$$

Rearrange to form a Quadratic Equation in $\sin y$

Check if the values of $x(=2\sin y)$ satisfy the given equation

Observe that $x>0$ as $-\dfrac\pi2<\arcsin(u)<0$ for $u<0$

lab bhattacharjee
  • 274,582